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Homework Help: Statics problem involving moments and reaction forces

  1. Mar 14, 2015 #1
    1. The problem statement, all variables and given/known data

    The right angle boom which supports the 230-kg cylinder is supported by three cables and a ball-and-socket joint at O attached to the vertical x-y surface. Determine the reactions at O and the cable tensions.
    2. Relevant equations
    = r x F
    unit vector = (vector)/(magnitude of vector)
    weight = weight of hanging mass = 230*9.81 = 2256.3 N
    3. The attempt at a solution
    So far, I've gotten two of the reaction forces and two of the cable tensions. Using unit vectors, I was able to solve for the tension vectors:
    T_ac = <-.4745, .4745, -.7414>T_ac
    T_bd = <0, .7880, -.6156>T_bd
    T_be = <0, 0, -1>T_be

    Because the moment about an axis sums to zero:
    ΣM_x = 0
    0 = (weight of mass - .4745*T_ac - .7880*T_bd)*radius from axis
    T_ac = (1/.4745)(2256.3 - .7880*T_bd)

    ΣM_z = 0
    0 = (T_bd*.7880 - .5*weight)*radius from axis
    T_bd = 1431.66 N which goes to 1430 N (The homework site I'm using only allows three significant digits.)

    T_ac = (1/.4745)(2256.3 - .7880*T_bd)
    T_ac = 2377.56 N which goes to 2380 N

    Up until this point, all the forces were equidistant from the axis in question.

    Somewhere in here my numbers get messed up - the previous two tension forces are correct, but this one is wrong.

    ΣM_y = 0
    0 = T_ac*-.4745*2.5 - T_be*-1*1.9 - T_be*1.9 + .6156*1.9*T_bd
    T_be = (1/1.9)(T_ac*.4745*2.5 - .6156*1.9*T_bd)
    T_be = 606 N

    Solving for the reaction forces:
    ΣF_x = 0
    0 = T_ac*-.4745 + O_x
    O_x = 1130 N

    ΣF_y = 0
    0 = O_y + .4745*T_ac + T_bd*.7880 - weight
    O_y = 0 N

    These two reaction forces are correct, so if I'm doing something wrong up to this point, I'm making compensating errors. The next reaction force is incorrect, but it depends on T_be and since I know that number is wrong, even if my method here is correct, I won't be able to get the right answer.

    ΣF_z = 0
    0 = -.7414*T_ac - .6156*T_bd - T_be + O_z
    O_z = 3250 N

    I'm not sure what I'm doing wrong, but at this point I'd be happy to find any mistake!

    Thanks for your time!
  2. jcsd
  3. Mar 14, 2015 #2


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    Just giving you work a quick eyeball, TBD should have three non-zero components.

    Also, when determining the sign of the components of a vector TBD, for example, these are usually calculated (xD - xB, yD - yB, zD - zB). [note the order]
  4. Mar 14, 2015 #3
    I'm not quite sure I follow you. I calculated the unit vector for T_bd as follows:

    B: <1.9, 0, 2.5>
    D: <1.9, 3.2, 0>

    BD = D - B
    = <0, 3.2, -2.5>
    n_bd = <0, 3.2, -2.5> / (3.2^2 + 2.5^2)^.5
    T_bd = n_bd * T_bd
    T_bd = <0, .7880, -.6156>

    Am I messing up the initial values for B and D?
  5. Mar 15, 2015 #4


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    No, it's my mistake here. I read the distance wrong from the diagram of the frame in the OP. :frown:
  6. Mar 15, 2015 #5


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    My calculations for T_ac and T_bd agree with yours. :smile:

    Why are you convinced that T_be is wrong?

    I agree with your calculation of T_be. What is O_z supposed to be if 3250 N is incorrect? :sorry:
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